Question

find the sum of first 40 integers divisible by 6

Answer 1

hey your answer is 4920


=6+12+18+24......240.
6(1+2+3+4+.....40)

{sum of 1to 10=55}
{sum of 11 to 20=155}
{sum of 21to 30=255}
{sum of 31to40=355}

so answer is 6(55+155+255+355)
6(820)=4920

Answer 2

6,12,18,24,30,36,42,48...240

This forms an Arithmetic Progression

a=First Term = 6

Common Difference=d= 6

$\boxed{\bold{Sum.of. n. terms=\frac{n}{2}(First Term+ Last Term) }}$

$\boxed{\bold{Sum.of. n. terms=\frac{40}{2}(6+240)=20*246=4920}}$